MATH 250: THE DISTRIBUTION OF PRIMES. ζ(s) = n s,

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1 MATH 50: THE DISTRIBUTION OF PRIMES ROBERT J. LEMKE OLIVER For s R, define the function ζs) by. Euler s work on rimes ζs) = which converges if s > and diverges if s. In fact, though we will not exloit this fact just yet, we may also make sense of ζs) for s C rovided that Rs) >. The connection between ζs) and rime numbers is the following. n= n s, Lemma. Euler roduct for ζs)). For Rs) >, we have.) ζs) = ). s rime Proof. Geometric series + unique factorization. Notation. Going forward, whenever we index a summation or roduct by a variable, we will imlicitly assume it is to be taken only over rime values of. This factorization affords another roof that there are infinitely many rimes. Proosition.. There are infinitely many rimes. Proof. Were there only finitely many rimes, exanding the right-hand side of.) as a geometric series would result in a series that converges absolutely at s =. However, the left-hand side diverges, therefore the right-hand side must as well, so there must be infintely many rimes. In fact, the roof of Proosition. yields rather more information: Theorem.3 The rimes are not too sarse. ). The series diverges. Proof. The roof of Proosition. shows that the infinite roduct /) diverges. Thus, ) = +, and alying log to both sides, we find log ) = + Date: Setember 6, 07.

2 ROBERT J. LEMKE OLIVER as well. Next, recall the Taylor series for log x), log x) = x + x + x , which is absolutely convergent if x <. We therefore find log ) = + + = + k. k We know that the left-hand side above diverges as N, so if we can show that the contribution on the right-hand side from the terms involving k with k is convergent, the theorem will follow. Naturally, it is now our goal to establish this convergence. We use a straightforward comarison test, noting that the summands are ositive. As the Taylor series for log x) is absolutely convergent for x < and / / for any rime, we have k k = =, k k k / where we have used the formula for the sum of a geometric series and the simle inequality /) for all rimes. The it as N of the term above is bounded by ζ), so it is convergent. Thus, we have log and taking the it as N yields the result. ) ζ), Remark. Theorem.3 rovides much more information than Euclid s roof of the infinitude of the rimes. For examle, it shows that at least tyically, the n-th rime n must satisfy n n, since the series /n converges. In fact, in the same wishy-washy, tyical sense, we must have n n +ɛ for any ɛ > 0, or even n nlog n) +ɛ, which is very close to what is rovided by the rime number theorem, that n n log n. More secifically, using techniques to be develoed later in this lecture, Theorem.3 establishes for examle that for any ɛ > 0.) su πx) = +. x/log x) +ɛ

3 EULER S WORK ON PRIMES 3 To contrast this with Euclid s roof, set a =, and for n define a n+ = a... a n +. Notice that for n, a n+ = a... a n )a n + = a n )a n + = a n a n +. In articular, a n+ is close to a n, and certainly a n+ a n. Just using this inequality, it follows that a n n, so that the same bound holds for n, namely n n. This holds unconditionally, and not in any wishy-washy sense, but all we may deduce from this is that which is markedly worse than.). inf πx) log log x > 0, Surrisingly, even though Theorem.3 is about showing that the rimes are not too sarse, very similar ideas show that the rimes cannot be too common. πx) Theorem.4 The rimes are not too common. ). x = 0. Proof. As a first and quite crude aroximation, note that every rime aart from is odd, and that there are x odd numbers x. Thus, πx) x + x )/ = x. To do better than this, we naturally want to incororate more divisibility conditions other than just those coming from two. However, in carrying this out, it becomes rather ainful rather quickly to come u with exact exressions akin to the x count of odd numbers u to x. Write out the corresonding exression for numbers that are relatively rime to 6. Do you have the stamina to do it for 30? For ?) We therefore forgo our fetish for exact formulae; instead, as with something like the rime number theorem, we write down an aroximate formula and kee track of how close it is. Let N be a fixed integer i.e., indeendent of x) and set P N) =. The roortion of integers n x that have no rime factor N is roughly /). In fact, this roortion is exact if x is an integral multile of P N), i.e. #{n < x : gcdn, P N)) = } = x if x = kp N) for some k Z. Since this count is exact at multiles of P N), between multiles of P N) the count could be off by at most the difference between the two endoint values, which we crudely bound by P N). Thus, for any x,.3) #{n < x : gcdn, P N)) = } = x ) + OP N)). Relating this back to πx), any rime is either at most N or relatively rime to N. There are πn) of the first tye, while those of the second are included in the count.3). Thus, πx) πn) + x ) + OP N)). Since N was fixed, this shows that πx).4) x ) + πn) + OP N)) x ), = ).

4 4 ROBERT J. LEMKE OLIVER From before, we know that whence ) = +, Inserting this into.4) yields the theorem. ) = 0. Remark. It is temting to use the ideas in the roof of Theorem.4 to attack a version of the rime number theorem. In articular, every integer n x greater than is either rime or admits a rime factor x /. This gives a quantitative version of the Sieve of Eratosthenes,.5) πx) = πx / ) + #{n x : gcdn, P x / )) = }. One might then hoe that the formula in.3) holds with some controlled error, something like.6) #{n x : gcdn, P x / )) = } = x ) + small error). x / However, it turns out an exression like this cannot hold with any lausible notion of small! Theorem.5 Mertens theorem). For any z, we have ) e γ log z, z where γ = is the Euler-Mascheroni constant. We will give a ost hoc roof of Theorem.5 in a later lecture after we ve roved the rime number theorem even though Theorem.5 redates the rime number theorem), but for right now, let us be content with observing that it imlies x / ) e γ log x. Thus, if.6) held, then.5) would yield πx) e γ x log x. This contradicts the rime number theorem, since e γ = 0.56 /, so something must be awry. The only room to give is in our assumtion of.6), so it must be the small error in.6) is not so small. Desite this being a failed and fundamentally flawed attemt at roving the rime number theorem, there is merit in the aroach. First, taking N = log x in the roof of Theorem.4 would yield su πx) x/ log log x <,

5 EULER S WORK ON PRIMES 5 and this choice of N is relatively easy to justify. In fact, with more effort than we care to ut in here, a careful treatment of the error term in.3) justifies a choice of N that shows su πx) x log log x/ log x <, which, combined with.), is starting to zero in on the rime number theorem. Second, there are roblems like the twin rime conjecture where we don t know whether there are infintely many such rimes, let alone whether an analogue of the rime number theorem holds. However, exanding on the ideas above, it is ossible to obtain uer bounds on the number of such rimes. This is the basis of the field called sieve theory. The first erson to develo these ideas was Viggo Brun. A very leasing consequence of Brun s work is the statement that the sum of the recirocals of the twin rimes converges. In light of the discussion around Theorem.3, this amounts to showing that there are not too many twin rimes.) We unfortunately won t send very much time on Brun s work in this course, though we will send some time discussing the so-called Selberg sieve when it comes time to rove the Maynard-Tao theorem.